A rod of length 3 m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at :
- 1.5 m
- 2 m
- 2.5 m
- 3.0 m
Answer
(2) 2 m
Solution
Mass per unit length: = kx
xcm = 0L x dx 0L dx = 0L kx2 dx 0L kx dx
= L3/3L2/2
= 2L3
xcm = 2 33 = 2 m
Answer: (2)